* * * {LAST QUESTION; MEDAL} * * * A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 15 hours. The population standard deviation is 2 hours. What is the 95% confidence interval for the mean? 14.88−15 hours 14.88−15.12 hours 15−15.12 hours 14.76−14.24 hours
I cant help but here is someone who might be able to: @hartnn
Do we have an Empirical Rule? You may wish to try two standard deviations on either side of the mean.
Yes, 68% at 1 and -1 95% at 2 and -2 99.7% at 3 and -3 But for 95% from the standard deviation, i got 11 and 19, which isn't one of the options..?
Have you considered that it wants the interval of the MEAN? You may wish to use \(SD_{Mean} = \dfrac{2}{\sqrt{1000}}\)
15 = 2/31.6227 What do I do with this?
How did you get 11-19? It's the same thing, just with the right Standard Deviation.
What is the right standard deviation? Sorry, I don't understand.
Did you miss \(SD_{Mean}\) that I supplied, above?
No, but I just don't know what to do with that. It would be 2^15?
Reset!!!! How did you find 11-19? EXACTLY what did you do to get that?
Add and subtract the SD from both sides of the mean.
What was I supposed to do?
Why not just do that again? You used \(SD_{Population} = 2\) Do it again with \(SD_{Mean} = \dfrac{2}{\sqrt{1000}}\)
oh, like add/subtract 30 from both sides?
No, the Standard Deviation of the Mean is \(\dfrac{2}{\sqrt{1000}} = 0.0632455532\) Where are you getting 30?
I multiplied 2 and 15. So on the right side, I add 0.063 and subtract that number on the left? From the mean?
Why are you doing ANYTHING with 2 after calculating the Standard Deviation of the Mean? The center of your confidence interval is 15. Never change that. The Standard deviation is 0.0632455532. Go!
Ok this is just confusing, sorry.....so, 15.0632, 15.1264, 15.1896 on the left side 14.9368, 14.8736, 14.8104 on the right?
Oh I see now. Thx for the help.
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